Contents

These Solutions are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Application of Integrals |

Exercise |
Ex 8.1, Ex 8.2 |

Number of Questions Solved |
20 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

### Chapter 8 Application of Integrals Exercise 8.1

**Question 1.**

Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4, and the x-axis.

**Solution:**

The curve y² = x is a parabola with vertex at origin.Axis of x is the line of symmetry, which is the axis of parabola. The area of the region bounded by the curve, x = 1, x=4 and the x-axis. Area LMQP

**Question 2.**

Find the area of the region bounded by y² = 9x, x = 2, x = 4 and x-axis in the first quadrant

**Solution:**

The given curve is y² = 9x, which is a parabola with vertex at (0, 0) and axis along x-axis. It is symmetrical about x-axis, as it contains only even powers of y. x = 2 and x = 4 are straight lines parallel toy-axis at a positive distance of 2 and 4 units from it respectively.

∴ Required area = Area ABCD

**Question 3.**

Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

**Solution:**

The given curve x² = 4y is a parabola with vertex at (0,0). Also since it contains only even powers of x,it is symmetrical about y-axis.y = 2 and y = 4 are straight lines parallel to x-axis at a positive distance of 2 and 4 from it respectively.

**Question 4.**

Find the area of the region bounded by the ellipse

**Solution:**

The equation of the ellipse is

The given ellipse is symmetrical about both axis as it contains only even powers of y and x.

**Question 5.**

Find the area of the region bounded by the ellipse

**Solution:**

It is an ellipse with centre (0,0) and length of major axis = 2a = 2×3 = 6 and length of minor axis = 2b = 2 × 2 = 4.

**Question 6.**

Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4. .

**Solution:**

Consider the two equations x² + y² = 4 … (i)

and x = √3y i.e. …(ii)

(1) x² + y² = 4 is a circle with centre O (0,0) and radius = 2.

**Question 7.**

Find the area of the smaller part of the circle x² + y² = a² cut off by the line

**Solution:**

The equation of the given curve are

x² + y² = a² …(i) and …(ii)

Clearly, (i) represent a circle and (ii) is the equation of a straight line parallel to y-axis at a

**Question 8.**

The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.

**Solution:**

Graph of the curve x = y² is a parabola as given in the figure. Its vertex is O and axis is x-axis. QR is the ordinate along x = 4

**Question 9.**

Find the area of the region bounded by the parabola y = x² and y = |x|.

**Solution:**

Clearly x² = y represents a parabola with vertex at (0, 0) positive direction of y-axis as its axis it opens upwards.

y = |x| i.e., y = x and y = -x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.

The required region is the shaded region as shown in the figure. Since both the curve are symmetrical about y-axis. So,

Required area = 2 (shaded area in the first quadrant)

**Question 10.**

Find the area bounded by the curve x² = 4y and the line x = 4y – 2

**Solution:**

Given curve is x² = 4y …(i)

which is an upward parabola with vertex at (0,0) and it is symmetrical about y-axis

Equation of the line is x = 4y – 2 …(ii)

Solving (i) and (ii) simultaneously, we get; (4y – 2)² = 4y

⇒ 16y² – 16y + 4 = 4y

⇒ 4y² – 5y + 1 = 0

**Question 11.**

Find the area of the region bounded by the curve y² = 4x and the line x = 3.

**Solution:**

The curve y² = 4x is a parabola as shown in the figure. Axis of the parabola is x-axis. The area of the region bounded by the curve y² = 4x and the line x = 3 is

A = Area of region OPQ = 2 (Area of the region OLQ)

Choose the correct answer in the following Exercises 12 and 13:

**Question 12.**

Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is

(a) π

(b)

(c)

(d)

**Solution:**

(a) x² + y² = 4. It is a circle at the centre (0,0) and r=2

**Question 13.**

Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is

(a) 2

(b)

(c)

(d)

**Solution:**

(b) y² = 4x is a parabola

### Chapter 8 Application of Integrals Exercise 8.2

**Question 1.**

Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.

**Solution:**

Area is bounded by the circle 4x² + 4y² = 9 and interior of the parabola x² = 4y.

Putting x² = 4y in x² + y² =

We get 4y + y² =

**Question 2.**

Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.

**Solution:**

Given circles are x² + y² = 1 …(i)

and (x – 1)² + y² = 1 …(ii)

Centre of (i) is O (0,0) and radius = 1

**Question 3.**

Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.

**Solution:**

Equation of the parabola is y = x² + 2 or x² = (y – 2)

Its vertex is (0,2) axis is y-axis.

Boundary lines are y = x, x = 0, x = 3.

Graphs of the curve and lines have been shown in the figure.

Area of the region PQRO = Area of the region OAQR-Area of region OAP

**Question 4.**

Using integration find the area of region bounded by the triangle whose vertices are (-1,0), (1,3) and (3,2).

**Solution:**

The points A (-1,0), B( 1,3) and C (3,2) are plotted and joined.

Area of ∆ABC = Area of ∆ ABL + Area of trap. BLMC – Area of ∆ACM …(i)

The equation of the line joining the points

**Question 5.**

Using integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4.

**Solution:**

The given lines are y = 2x + 1 …(i)

y = 3x + 1 …(ii)

x = 4 …(iii)

Subtract (i) from eq (ii) we get x = 0, Putting x = 0 in eq(i) y = 1

∴ Lines (ii) and (i) intersect at A (0,1) putting x = 4 in eq. (2) =>y = 12 + 1 = 13

The lines (ii) and (iii) intersect at B (4,13) putting x=4ineq. (i):y = 8 + 1 = 9

∴ Lines (i) and (ii); Intersect at C (4,9),

**Question 6.**

Smaller area bounded by the circle x² + y² = 4 and the line x + y = 2

(a) 2 (π – 2)

(b) π – 2

(c) 2π – 1

(d) 2(π + 2)

**Solution:**

(b) A circle of radius 2 and centre at O is drawn.The line AB: x + y = 2 is passed through (2,0) and (0,2). Area of the region ACB

= Area of quadrant OAB – Area of ∆OAB …(i)

**Question 7.**

Area lying between the curves y² = 4x and y = 2x.

(a)

(b)

(c)

(d)

**Solution:**

(b) The curve is y² = 4x …(1)

and the line is y = 2x …(2)

We hope the NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals help you. If you have any query regarding NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals, drop a comment below and we will get back to you at the earliest.

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